Find the Parabola Through (0,40) with Vertex (1,45) (1,45) , (0,40)

$(1, 45)$ , $(0, 40)$

Step 1

The general equation of a parabola with vertex $(h, k)$ is $y = a {(x - h)}^{2} + k$ . In this case we have $(1, 45)$ as the vertex $(h, k)$ and $(0, 40)$ is a point $(x, y)$ on the parabola. To find $a$ , substitute the two points in $y = a {(x - h)}^{2} + k$ .

$40 = a {(0 - (1))}^{2} + 45$

Step 2

Using $40 = a {(0 - (1))}^{2} + 45$ to solve for $a$ , $a = - 5$ .

Rewrite the equation as $a {(0 - (1))}^{2} + 45 = 40$ .

$a {(0 - (1))}^{2} + 45 = 40$

Simplify $a {(0 - (1))}^{2} + 45$ .

Subtract $1$ from $0$ .

$a {(- (1))}^{2} + 45 = 40$

Simplify each term.

Multiply $- 1$ by $1$ .

$a {(- 1)}^{2} + 45 = 40$

Raise $- 1$ to the power of $2$ .

$a \cdot 1 + 45 = 40$

Multiply $a$ by $1$ .

$a + 45 = 40$

Move all terms not containing $a$ to the right side of the equation.

Subtract $45$ from both sides of the equation.

$a = 40 - 45$

Subtract $45$ from $40$ .

$a = - 5$

Step 3

Using $y = a {(x - h)}^{2} + k$ , the general equation of the parabola with the vertex $(1, 45)$ and $a = - 5$ is $y = (- 5) {(x - (1))}^{2} + 45$ .

$y = (- 5) {(x - (1))}^{2} + 45$

Step 4

Solve $y = (- 5) {(x - (1))}^{2} + 45$ for $y$ .

Remove parentheses.

$y = (- 5) {(x - (1))}^{2} + 45$

Multiply $- 5$ by ${(x - (1))}^{2}$ .

$y = - 5 {(x - (1))}^{2} + 45$

Remove parentheses.

$y = (- 5) {(x - (1))}^{2} + 45$

Simplify $(- 5) {(x - (1))}^{2} + 45$ .

Simplify each term.

Multiply $- 1$ by $1$ .

$y = - 5 {(x - 1)}^{2} + 45$

Rewrite ${(x - 1)}^{2}$ as $(x - 1) (x - 1)$ .

$y = - 5 ((x - 1) (x - 1)) + 45$

Expand $(x - 1) (x - 1)$ using the FOIL Method.

Apply the distributive property.

$y = - 5 (x (x - 1) - 1 (x - 1)) + 45$

Apply the distributive property.

$y = - 5 (x \cdot x + x \cdot - 1 - 1 (x - 1)) + 45$

Apply the distributive property.

$y = - 5 (x \cdot x + x \cdot - 1 - 1 x - 1 \cdot - 1) + 45$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y = - 5 (x^{2} + x \cdot - 1 - 1 x - 1 \cdot - 1) + 45$

Move $- 1$ to the left of $x$ .

$y = - 5 (x^{2} - 1 \cdot x - 1 x - 1 \cdot - 1) + 45$

Rewrite $- 1 x$ as $- x$ .

$y = - 5 (x^{2} - x - 1 x - 1 \cdot - 1) + 45$

Rewrite $- 1 x$ as $- x$ .

$y = - 5 (x^{2} - x - x - 1 \cdot - 1) + 45$

Multiply $- 1$ by $- 1$ .

$y = - 5 (x^{2} - x - x + 1) + 45$

Subtract $x$ from $- x$ .

$y = - 5 (x^{2} - 2 x + 1) + 45$

Apply the distributive property.

$y = - 5 x^{2} - 5 (- 2 x) - 5 \cdot 1 + 45$

Simplify.

Multiply $- 2$ by $- 5$ .

$y = - 5 x^{2} + 10 x - 5 \cdot 1 + 45$

Multiply $- 5$ by $1$ .

$y = - 5 x^{2} + 10 x - 5 + 45$

Add $- 5$ and $45$ .

$y = - 5 x^{2} + 10 x + 40$

Step 5

The standard form and vertex form are as follows.

Standard Form: $y = - 5 x^{2} + 10 x + 40$

Vertex Form: $y = (- 5) {(x - (1))}^{2} + 45$

Step 6

Simplify the standard form.

Standard Form: $y = - 5 x^{2} + 10 x + 40$

Vertex Form: $y = - 5 {(x - 1)}^{2} + 45$

Step 7

Find the Parabola Through (0,40) with Vertex (1,45) (1,45) , (0,40)

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